CHEMISTRY Midterm #2 answer key October 25, 2005
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1 CHEMISTRY Mdterm #2 answer key October 25, 2005 Statstcs: Average: 70 pts (70%); Hghest: 97 pts (97%); Lowest: 33 pts (33%) Number of students performng at or above average: 62 (63%) Number of students performng at or below 55%: 24 (24%) PART I: MULTIPLE CHOICE 1. (2 pts) Whch compound wll not dssolve n water n large amounts: a. Mg(ClO 4 ) 2 ; b. Cu(NO 3 ) 2 ; c. (NH 4 ) 2 SO 4 ; d. Fe(OH) 2 e. NSO 4 ; f. CoCl 2 ; 2. (2 pts) Whch statement about strong acds s true? a. Strong acds are weak electrolytes. b. Strong acds are very concentrated. c. Strong acds are almost entrely converted to ons when dssolved n water. d. Acetc acd s a strong acd. e. All of the above are true. 3. (2 pts) Whch of the followng s a non-electrolyte? a. HBr b. Mg(OH) 2 c. NaCl 4. (2 pts) What s the correct formula for the hydronum on? a. H 2 O; b. H 3 O + ; c. H 3 O - ; d. C 2 H 6 e. KI d. OH - ; e. H - ; f. NH + 4 ; 5. (2 pts) Whch of the followng factors cause exchange reactons to occur? a. Formaton of a gas only b. Formaton of a precptate only c. Formaton of water only 6. (2 pts) What s the oxdaton number of Cl n Cl 2 O 5? a. +2 b. -2 c (2 pts) What s the oxdaton number of P n PO 4 3-? a. -3 b. +3 c (2 pts) What s the oxdaton number of H n MgH 2? a. -1 b. +1 c (2 pts) What s the oxdaton number of F n F 2? a. -1 b. +1 d. Formaton of a gas or a precptate e. Formaton of a gas, precptate or water d. -5 e. +5 d. +5 e. 5 d. +2 e. 2 c. 0 d. +2
2 e (2 pts) Whch dsplacement reacton wll not occur? a. Mg + CaSO 4 MgSO 4 + Ca b. Ba + 2 HCl BaCl 2 + H 2 c. 2 Al + 3 SnCl 2 2 AlCl Sn d. H AgNO 3 2 HNO Ag e. Mg + CuSO 4 Cu + MgSO (2 pts) Whch of the methods descrbed below wll yeld 100 ml of a M NaOH soluton? a. Add exactly 100 ml of water to 4.00 g of NaOH. b. Dssolve g NaOH n water and dlute to exactly 100 ml. c. Add exactly 100 ml of water to NaOH. d. Dssolve 4.00 g of NaOH n water and dlute to exactly 100 ml. e. Dlute 50.0 ml of 1.00 M NaOH to exactly 100 ml. 12. (2 pts) Whch of the followng s not an example of knetc energy? a. The moton of a molecule b. The moton of a golf ball c. The vbraton of an object d. A loosely held brck on the top of a buldng e. The moton of electrons through a wre 13. (2 pts) Determne the ncorrect relatonshp gven below. a cal/g = 312 J/g b cal = 1 kcal c kj = 1.05 x 10 4 cal d. 1 µj = 1 x 0-6 J e J = 1 kj 14. (2 pts) Whch substance has the hghest molar heat capacty? a. Alumnum (specfc heat = 0.92 J/g o C) b. Copper (specfc heat = 0.39 J/g o C) c. Iron (specfc heat = 0.46 J/g o C) d. Slver (specfc heat = 0.23 J/g o C) e. Lead (specfc heat = 0.13 J/g o C) 15. (2 pts) Whch statement s true? a. A postve change n enthalpy occurs wth endothermc processes. b. A postve change n enthalpy occurs wth exothermc processes. c. A postve change n enthalpy occurs when work s done on the surroundngs. d. A postve change n enthalpy occurs when work s done on the system. e. A postve change n enthalpy occurs when a process s endothermc and does work on the surroundngs. 16. (2 pts) Whch statement about the reacton below s ncorrect? H 2 O (l) H 2 O (g) H = kj a. The value for the gas to lqud value has the reverse sgn. b. The value for the sold to lqud transformaton s the same. c. The value for evaporatng 2 moles of water s 88.0 kj. d. The reacton s endothermc. e. The above data does not gve any nformaton about the sold to lqud transformaton.
3 PART II: SHORT ANSWER (Each short answer queston has a 1-pont value!!) 17. Molarty s a unt of soluton concentraton expressed n moles per lter. 18. When an element s oxdzed, t loses electrons. 19. A neutralzaton reacton nvolves the reacton of a(n) acd wth a(n) base. 20. A chemcal equaton wthout the spectator ons s called a net onc equaton. 21. The quantty of energy requred to ncrease the temperature of one gram of a sample by 1 C s called the specfc heat capacty. 22. Enthalpy change s equal to heat transfer at constant pressure. PART III: CHEMICAL REACTIONS 23. (20 pts) Wrte (clearly and dstnctly) the complete and net onc equatons for each of the followng reactons IN WATER. If no reacton occurs, wrte NR. a. ZnCl 2 + Na 2 SO 4 NO Reacton, All potental products are soluble onc compounds b. CuSO 4 + NaOH Cu 2+ (aq) + SO 4 2- (aq) + 2 Na + (aq) + 2 OH - (aq) 2 Na + (aq) + SO 4 2- (aq) + Cu(OH) 2 (s) Net: Cu 2+ (aq) + 2 OH - (aq) Cu(OH) 2 (s) c. AgNO 3 + Na 3 PO 4 3 Ag + (aq) + 3 NO 3 - (aq) + 3 Na + (aq) + PO 4 3- (aq) 3 NO 3 - (aq) + 3 Na + (aq) + Ag 3 PO 4 (s) Net: 3 Ag + (aq) + PO 4 3- (aq) Ag 3 PO 4 (s) d. CH 3 COOH + KOH CH 3 COOH (aq) + Na + (aq) + OH - (aq) CH 3 COO - (aq) + Na + (aq) + H 2 O (l) Net: CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) e. HBr + FeS FeS (s) + 2 H + (aq) + 2 Br - (aq) Net: FeS (s) + 2 H + (aq) Fe 2+ (aq) + 2 Br - (aq) + H 2 S (g) Fe 2+ (aq) + H 2 S (g)
4 24. (12 pts) Fnd whch of the equatons below represent redox reactons. Indcate the element that s reduced and the element that s oxdzed. Wrte ther oxdaton numbers before and after the reacton. a. NaNO 3 (s) + Pb (s) NaNO 2 (s) + PbO (s) redox reacton Oxdzed: Pb; oxdaton number changes from 0 to +2 Reduced: N; oxdaton number changes from +5 to +4 b. Na 2 SO 4 (s) + 4 C (s) Na 2 S (s) + 4 CO (s) redox reacton Oxdzed: C; oxdaton number changes from 0 to +2 Reduced: S; oxdaton number changes from +6 to -2 c. MnO 2 (s) + 4 HCl (aq) MnCl 2 (aq) + Cl 2 (g) + 2 H 2 O (l) redox reacton Oxdzed: Cl; oxdaton number changes from -2 to 0 Reduced: Mn; oxdaton number changes from +4 to +2 d. SO 2 (s) + Na 2 O (s) Na 2 SO 3 (s) NOT a redox reacton 25. (9 pts) For each of the followng dsplacement reactons, predct whether the process wll occur from left to rght or from rght to left. Soluton: Decsons are made based on the actvty seres. A more actve metal (closer to the top of the lst) wll always dsplace a less actve one from ts compounds. a. 3 Mg 2+ (aq) + 2 Cr (s) or 2 Cr 3+ (aq) + 3 Mg (s) to the left b. 2 H + (aq) + 2 L (s) or 2 L + (aq) + H 2 (g) to the rght c. 2 Ag+ (aq) + Fe (s) or Fe 2+ (aq) + 2 Ag (s) to the rght PART IV: CONCEPTS 26. (3 pts) Below s gven a process that results n a change of both volume and temperature. a. Has any work been done? If so, s the work postve or negatve? There s a change n volume so work was done. Snce the fnal volume s greater than the ntal, w > 0. b. Has there been an enthalpy change? If so, s the reacton exothermc or endothermc?
5 There s a change of temeperature, so heat transfer occurred as well. Snce T fnal > T ntal, q > 0. Hence the process s endothermc. 27. (3 pts) A reacton s carred out n a cylnder, ftted wth a movable pston (see below). The startng volume s 5.00 L. At constant pressure of 1 atm the changes n enthalpy and energy are: H = kj and E = kj. Does the volume ncrease, decrease or reman the same? Soluton: We know that E = H + w = H + p V = H + p (V fnal V ntal ). Based on the values of E and H, one concludes that p V > 0, meanng that V fnal > V ntal. Hence the process occurred wth a volume ncrease. PART V: CALCULATION PROBLEMS (Show your work n ts entrety. Do not provde just a sngle number!). 28. (5 pts) Determne the standard enthalpy change for the process: C 2 H 4 (g) + 6 HCl (g) 2 CHCl 3 (g) + 4 H 2 (g) usng the nformaton gven below: 2 C (s) + 2 H 2 (g) C 2 H 4 (g) H o = 52.3 kj H 2 (g) + Cl 2 (g) 2 HCl (g) H o = kj C (s) + ½ H 2 (g) + 3/2 Cl 2 (g) CHCl 3 (g) H o = kj Soluton: We begn by dentfyng reactants and products n the equaton n queston and als o where they appear n the set of equatons. The followng changes/manpulatons are necessary: Equaton 1: Reverse: C 2 H 4 (g) 2 C (s) + 2 H 2 (g) H o = kj Equaton 2: Reverse and multply by 3: 6 HCl (g) 3 H 2 (g) + 3 Cl 2 (g) H o = kj Equaton 3: Multply by 2: 2 C (s) + H 2 (g) + 3 Cl 2 (g) 2 CHCl 3 (g) H o = kj Inspecton now shows that addng the three new equatons gves precsely the process under study. Therefore the summng of the enthalpy changes must gve us the enthalpy change for the above process: DH o = = kj 29. (5 pts) The standard molar enthalpes of formaton for several substances are gven below: FeS 2 (s) = kj/mol FeCl 2 (s) = kj/mol FeCl 3 (s) = kj/mol HCl (g) = kj/mol HCl (aq) = kj/mol H 2 S (g) = kj/mol H 2 O (g) = kj/mol Calculate H o for the reacton below:
6 2 FeS 2 (s) + 8 HCl (g) 2 FeCl 3 (s) + 4 H 2 S (g) + Cl 2 (g) Soluton: The enthalpy change for the reacton can be found by applyng the general formula: H o = Σ [# mol x H o f] products Σ [# mol x H o f] reactants Therefore: H o = [2 x (-399.5) + 4 x (- 20.6)] [2 x ( ) + 8 x (- 92.3)] = kj 30. (5 pts) When g of KNO 3 was dssolved n water n a calormeter, the temperature fell from o C to o C. The heat capacty of the calormeter s 682 J/ o C. Wrte the complete thermochemcal equaton for the process: KNO 3 (s) K + (aq) + NO 3 - (aq) Soluton: It s based on the fact that q system = - q surroundngs q surroundngs = (heat capacty) calormeter x T = 682 x ( o C) = J Therefore q system = J Ths s the amount of heat absorbed by g x 1 mol/ g = mol of KNO 3 Per mole: 7407 J/ mol = J = kj DH o = kj 31. (4 pts) BONUS PROBLEM (In order to receve credt, you have to solve the problem completely!). A mxture was made form the followng: 25.0 g of water at 15.0 o C, 45.0 g of water at 50.0 o C, and 15.0 g of water at 37.0 o C. What s the fnal temperature of the mxture? Soluton: It s based on the fact that: q 1 + q 2 + q 3 = 0, due to the law of conservaton of energy c x m 1 x ( T) 1 + c x m 2 x ( T) 2 + c x m 3 x ( T) 3 = 0 m 1 x ( T) 1 + m 2 x ( T) 2 + m 3 x ( T) 3 = 0 m 1 x (T f T 1 ) + m 2 x (T f T 2 ) + m 3 x (T f T 3 ) = 0 (m 1 + m 2 + m 3 ) x T f = m 1 x T 1 + m 2 x T 2 + m 3 x T 3 T f = (m 1 x T 1 + m 2 x T 2 + m 3 x T 3 ) / (m 1 + m 2 + m 3 ) T f = (25.0 x x x 37.0) / ( ) = 37.4 o C T f = 37.4 o C
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